cven214- lecture 1 stress -dr. wael alnahhal

COLLEGE OF ENGINEERING DEPARTMENT OF CIVIL & ARCHITECTURAL ENGINEERING

CVEN 214: STRENGTH OF MATERIALS

WAEL I. ALNAHHAL, Ph. D., P. Eng

Spring, 2015

Chapter 1: STRESS

Introduction Mechanics of materials is a study of the relationship

between the external loads on a body and the intensity of the internal loads within the body.

This subject also involves the deformations and stability of a body when subjected to external forces.

Hence, this course will basically deal with loads, deformations and the forces acting on the materials

Definition: Mechanics of materials or strength of materials is a branch of applied mechanics that deals with the behaviour of solid bodies subjected to various types of loading

Compression Tension (stretched) Bending Torsion (twisted) Shearing

Objectives

Equilibrium of a Deformable Body External Forces 1.Surface Forces - Caused by direct

contact of other bodys surface

-Either concentrated or distributed

2.Body Forces

- other body exerts a force without contact (gravitational force)

Equilibrium of a Deformable Body Support Reactions Surface forces developed at the supports/points of

contact between bodies.

Equilibrium of a Deformable Body

Equations of Equilibrium Equilibrium of a body requires a balance

of forces and a balance of moments

For a body with x, y, z coordinate system with origin O,

Best way to account for these forces is to draw the bodys Free-Body Diagram (FBD).

0M 0F == O

0 , 0 , 0

0 , 0 , 0

===

===

zyx

zyx

MMM

FFF

Free Body Diagram (FBD) A free body diagram is a graphic, symbolic

representation of a body with all connecting physical "parts" removed and replaced by the applied external forces and/or moments.

Loads, distributed or concentrated, are removed and replaced with representational force system

Free Body Diagram (FBD)

Equilibrium of a Deformable Body Internal Resultant Loadings Objective of FBD is to determine the resultant force

and moment acting within a body. In general, there are 4 different types of resultant

loadings: a) Normal force, N b) Shear force, V c) Torsional moment or torque, T d) Bending moment, M

Coplanar Loadings

, 0

, 0

, 0

x

y

o x

for N F

for V F

for M M

=

=

=

Free Body Diagram

Understand the concepts of normal and shear stress Analyze and design of members subjected on axial load

or shear

Introduction

Applied External Load

Internal Stresses Deformations

Stability and Deformations of a body depend on the intensity of the external forces applied on the body as well as the load carrying capacity of the body

Determine Using Equilibrium Equations

Concept of Stress

Obtaining the distribution of internal loading (or force) is of primary important in mechanics of materials.

To solve this problem its is necessary to establish the concept of stress

Stress is the intensity of internal force at a specific plane (area) passing through a point.

(a) Tensile (b) Compressive and (c) Shear Loading

Types of externally applied Loading

Concept Stress

Types of Stress Stress is the intensity of internal force at a specific plane

(area) passing through a point. 1. Normal Stress, The internal force per unit area acting normal to A

2. Shear Stress, The intensity of force per unit area acting on a plane Tangent to A

2

0lim / ( )zz A

FF N m or pascal PaA A

= =

2

0

2

0

lim ( )

lim ( )

xzx A

yzy A

FF Nm or pascal PaA A

FF Nm or pascal PaA A

= =

= =

Average Normal Stress in an Axially Loaded Bar When a cross-sectional area bar (long & slender

mechanical or structural element) is subjected to axial force through the centroid, it is only subjected to normal stress.

Stress is assumed to be averaged over the area.

Average Normal Stress in an Axially Loaded Bar

Average Normal Stress Distribution When an axially loaded uniform bar is subjected to a constant uniform deformation,

APAP

dAdFA

=

=

=

= average normal stress P = resultant normal force A = cross sectional area of bar

Types of Normal Stress For Vertical Equilibrium of Axially loaded bars The two normal stress components Must be equal in

magnitude but opposite in direction

For a bar subjected to several external loads along its axis or a change in its cross-sectional area, as result , the could be different from one area to another on the bar.

Determination of the maximum P/A= is very important in this case

Maximum Average Stress

Procedure for Normal Stress Analysis

Generally, the equation gives the average normal stress due to internal force P

For axially loaded bars Internal Loading determinations: Section the members perpendicular to its longitudinal

axis at the point of interest Use the necessary FBD and equilibrium equations to

determine the internal axial force P Average Normal Stress determination: Determine the A and compute the

PA

=

PA

=

Example 1.6 The bar has a constant width of 35 mm and a thickness of 10 mm.

Determine the maximum average normal stress in the bar when it is

subjected to the loading shown.

Solution: Example 1.6

By inspection, different sections have different internal forces.

Graphically, the normal force diagram is as shown.


By inspection, the largest loading is in region BC,

kN 30=BCPSince the cross-sectional area of the bar is constant, the largest average normal stress is

( )( )( )

36 2

30 1085.7 10 N/m or 85.7MPa (Ans)

0.035 0.01BC

BCP xA

= = =

Example 1.7

3kN/m 80=st

Example 1.8 The casting is made of steel that has a specific weight of . Determine the average compressive stress acting at points A and B.


By drawing a free-body diagram of the top segment, the internal axial force P at the section is

( )( ) ( )kN 042.8

02.08.080

0 ;02

==

==+

PP

WPF stz

The average compressive stress becomes

( )(Ans) kN/m 0.64

2.0042.8 2

2 ===

AP

Example 1.9

Example 1.9: Solution

Average Shear Stress Shear Stress:The intensity of force per unit area acting on a plane Tangent to A The average shear stress distributed over each

sectioned area that develops a shear force.

AV

avg = = average shear stress at section V = internal resultant shear force at the section determine from equilibrium equations A = area at that section

Two Different types of shear a) Single Shear connections: e.g lap joints

b) Double Shear-surfaces: eg double lap-joints

Shear Equilibrium

Procedure for Shear Stress Analysis

Generally, the equation is used to compute the average shear stress in materials via the steps below:

Internal Shear Stress determinations: Section the member at the point of interest Draw the necessary FBD and use the equilibrium

equations to determine the internal shear force V Average Normal Stress determination: Determine the A and compute the

VA

=

VA

=

Example 1.10

Example 1.12 The inclined member is subjected to a compressive force of 3000 N. Determine the average compressive stress along the smooth areas of contact defined by AB and BC, and the average shear stress along the horizontal plane defined by EDB.

Solution:

The compressive forces acting on the areas of contact are

( )( ) N 240003000 ;0

N 180003000 ;0

54

53

===+

===+

BCBCy

ABABx

FFF

FFF

The shear force acting on the sectioned horizontal plane EDB is


N 1800 ;0 ==+ VFx

Average compressive stresses along the AB and BC planes are

( )( )

( )( ) (Ans) N/mm 20.140502400

(Ans) N/mm 80.14025

1800

2

2

==

==

BC

AB

( )( ) (Ans) N/mm 60.040751800 2==avg

Average shear stress acting on the BD plane is

Allowable Stress Many unknown factors that influence the actual

stress in a member. A factor of safety is needed to obtained allowable

load. The factor of safety (F.S.) is a ratio of the failure

load divided by the allowable load

allow

fail

allow

fail

allow

fail

SF

SF

FF

SF

=

=

=

.

.

.

Example 1.14 The control arm is subjected to the loading. Determine to the nearest 5 mm the required diameter of the steel pin at C if the allowable shear stress for the steel is . Note in the figure that the pin is subjected to double shear. MPa 55=allowable


( ) ( ) ( )( )( )( ) kN 3002515 ;0

kN 502515 ;0

kN 150125.025075.0152.0 ;0

53

54

53

===+

==+=+

====+

yyy

xxx

ABABC

CCF

CCF

FFM

For equilibrium we have


The pin at C resists the resultant force at C. Therefore,

( ) ( ) kN 41.30305 22 =+=CF

mm 8.18

mm 45.2462

m 1045.2761055

205.15

2

263

2

=

=

=

==

d

d

VAallowable

The pin is subjected to double shear, a shear force of 15.205 kN acts over its cross-sectional area between the arm and each supporting leaf for the pin. The required area is

Use a pin with a diameter of d = 20 mm. (Ans)

Example 1.17 The rigid bar AB supported by a steel rod AC having a diameter of 20 mm and an

aluminum block having a cross sectional area of 1800 mm2. The 18-mm-diameter

pins at A and C are subjected to single shear. If the failure stress for the steel and

aluminum is a & respectively, and the failure

shear stress for each pin is , determine the largest load P that can

be applied to the bar. Apply a factor of safety of F.S. = 2.

( ) MPa 680=failst ( ) MPa 70=failalMPa 900=fail

Solution: Example 1.17 The allowable stresses are

( )( )

( )( )

MPa 4502

900..

MPa 352

70..

MPa 3402

680..

===

===

===

SF

SF

SF

failallow

failalallowal

failstallowst

There are three unknowns and we apply the equations of equilibrium,

( ) ( )( ) ( ) (2) 075.02 ;0

(1) 0225.1 ;0

==+

==+

PFM

FPM

BA

ACB


We will now determine each value of P that creates the allowable stress in the rod, block, and pins, respectively.

For rod AC, ( ) ( ) ( ) ( )[ ] kN 8.10601.010340 26 === ACallowstAC AF

Using Eq. 1, ( )( ) kN 17125.1

28.106==P

For block B, ( ) ( ) ( )[ ] kN 0.631018001035 66 === BallowalB AF

Using Eq. 2, ( )( ) kN 168

75.020.63

==P


For pin A or C, ( ) ( )26450 10 0.009 114.5 kNAC allowV F A = = = = Using Eq. 1, ( )( ) kN 183

25.125.114

==P

When P reaches its smallest value (168 kN), it develops the allowable normal stress in the aluminium block. Hence,

(Ans) kN 168=P

READING QUIZ

1. What is the normal stress in the bar if P=10 kN and 500mm?

a) 0.02 kPa

b) 20 Pa

c) 20 kPa

d) 200 N/mm

e) 20 MPa

READING QUIZ (cont)

2. What is the average shear stress in the internal vertical surface AB (or CD), if F=20kN, and AAB=ACD=1000mm?

a) 20 N/mm

b) 10 N/mm

c) 10 kPa

d) 200 kN/m

e) 20 MPa

CONCEPT QUIZ

1) The thrust bearing is subjected to the loads as shown. Determine the order of average normal stress developed on cross section through BC and D.

a) C > B > D

b) C > D > B

c) B > C > D

d) D > B > C

Slide Number 1IntroductionSlide Number 3ObjectivesObjectivesEngineering Statics ReviewEquilibrium of a Deformable BodyEquilibrium of a Deformable BodyEquilibrium of a Deformable BodyFree Body Diagram (FBD)Free Body Diagram (FBD)Equilibrium of a Deformable BodyCoplanar LoadingsReview ProblemsSlide Number 15Slide Number 16Slide Number 17IntroductionConcept of StressConcept StressTypes of StressAverage Normal Stress in an Axially Loaded BarAverage Normal Stress in an Axially Loaded BarTypes of Normal StressSlide Number 25Procedure for Normal Stress AnalysisExample 1.6Solution: Example 1.6Slide Number 29Example 1.7Example 1.7Example 1.8Slide Number 33Example 1.9Example 1.9: SolutionAverage Shear StressTwo Different types of shearShear EquilibriumProcedure for Shear Stress AnalysisExample 1.10Example 1.10Slide Number 42Slide Number 43Example 1.12Slide Number 45Allowable StressExample 1.14Solution: Example 1.14Slide Number 49Example 1.17Slide Number 51Slide Number 52Slide Number 53 READING QUIZ READING QUIZ (cont) CONCEPT QUIZ

cven214- lecture 1 stress -dr. wael alnahhal

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